bounded convergence theorem

The bounded convergence theorem is a special case of the dominated convergence theorem, though it can be proved using slightly different techniques.

Theorem

For a measure space $(X,\mathcal{S},\mu )$ with $\mu <\infty$ (very important!), suppose there is a sequence of measurable functions $f_1,f_2,\cdots :X\to \mathbb{R}$ that converges pointwise to a function $f:X\to \mathbb{R}$. If there exits $c\in (0,\infty )$ such that

$$|f_k(x)|\le c\forall k\in {\mathbb{Z}}^{+}\forall x\in X$$

then

$$\underset{k\to \infty }{\lim }\int f_{k}d\mu =\int \underset{k\to \infty }{\lim }{f}_{k}d\mu =\int fd\mu$$

Basically, if your function is on a finite measure space, and converges pointwise to a bounded function then the limit and the integral can be exchanged. (The integral doesn’t leak).

Proof

This proof relies on Egorov’s Theorem, so we restrict the domain down to get uniform convergence while only cutting out an arbitrarily small set.

For $ϵ>0$ By Egorov’s theorem, $\exists E\subset X$ such that $\mu (X\backslash E)<\frac{ϵ}{4c}$ and $f_k\to f$ converges uniformly on $E$.

Using this set, we split the integral along this set

$$\begin{array}{rl}\left|\int f_{k}d\mu -\int fd\mu \right|& =\left|{\int }_{X\backslash E}{f}_{k}d\mu -{\int }_{x\backslash E}fd\mu +{\int }_E{f}_{k}d\mu -{\int }_{E}fd\mu \right|\\ & \le {\int }_{x\backslash E}|f_k|d\mu +{\int }_{X\backslash E}|f|d\mu +{\int }_E|f_k-f|d\mu \\ & \le \mu (X\backslash E)(\sup |f_k|+\sup |f|)+\mu (E)\underset{E}{\sup }|f_k-f|\\ & \le \frac{ϵ}{4c}(2c)+\mu (E)\underset{E}{\sup }|f_k-f|<ϵ\end{array}$$

The inequality on the last line holds because $f_k\to f$ uniformly so for k large, we can find the ${\sup }_E|f_k-f|<\frac{ϵ}{2\mu (E)}$ for all $x\in E$

Thus, the limit doesn’t leak and the limit can be interchange with the integral.